Showing posts from August, 2013

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In the excellent (and highly recommended) book "Fifty Challenging Problems in Probability with Solution", Frederick Mosteller poses "The Three-Cornered Duel":

A, B and C are to fight a three-cornered pistol duel. All know that A's chance of hitting his target is 0.3, C's is 0.5, and B never misses. They are to fire at their choice of target in succession in the order A, B, C, cyclically (but a hit man loses further turns and is no longer shot at) until only one man is left unit. What should A's strategy be? This is problem 20 in Mosteller's book, and it also appears (with an almost identical solution) in Larsen & Marx "An Introduction to Probability and Its Applications".

Mosteller's solution:

A is naturally not feeling cheery about this enterprise. Having the first shot he sees that, if he hits C, B will then surely hit him, and so he is not going to shoot at C. If he shoots at B and misses him, then B clearly shoots the more dan…

A, B and C are to fight a three-cornered pistol duel. All know that A's chance of hitting his target is 0.3, C's is 0.5, and B never misses. They are to fire at their choice of target in succession in the order A, B, C, cyclically (but a hit man loses further turns and is no longer shot at) until only one man is left unit. What should A's strategy be? This is problem 20 in Mosteller's book, and it also appears (with an almost identical solution) in Larsen & Marx "An Introduction to Probability and Its Applications".

Mosteller's solution:

A is naturally not feeling cheery about this enterprise. Having the first shot he sees that, if he hits C, B will then surely hit him, and so he is not going to shoot at C. If he shoots at B and misses him, then B clearly shoots the more dan…

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A bar bet as presented in the YouTube video The HARDEST Puzzle Yet! involves starting with three of the same number from 0 through 9, then adding mathematical operations that result in an evaluation of 6. For example, if we start with three of the number 6, one solution could be \[ 6+6-6=6 .\] I'll demonstrate a method for solving this bar bet puzzle starting with three of any number, say \( N \), which involves using at most three different mathematical operations (although some of these may be used many, many times).

If \( 0 \leq N \leq 2 \) we have the solutions

\begin{eqnarray}

(0! + 0! + 0!)! &= 6 \\

(1! + 1! + 1!)! &= 6 \\

2+2+2 &= 6.

\end{eqnarray} If \( N\geq 3\), concatenate the three numbers. Repeatedly applying the square-root operation we'll eventually end up with a result \( x \) with \( 3 \leq x < 9\). If we now take the greatest integer \(\lfloor x \rfloor\) we have an integer \( n \) with \( 3 \leq n \leq 8 \). If we can exhibit solutions for ea…

If \( 0 \leq N \leq 2 \) we have the solutions

\begin{eqnarray}

(0! + 0! + 0!)! &= 6 \\

(1! + 1! + 1!)! &= 6 \\

2+2+2 &= 6.

\end{eqnarray} If \( N\geq 3\), concatenate the three numbers. Repeatedly applying the square-root operation we'll eventually end up with a result \( x \) with \( 3 \leq x < 9\). If we now take the greatest integer \(\lfloor x \rfloor\) we have an integer \( n \) with \( 3 \leq n \leq 8 \). If we can exhibit solutions for ea…

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